If the points (2, 0), (0, 1), (4, 5) and (0, | vedclass

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(C) Let the equation of the circle passing through $(2, 0), (0, 1)$ and $(4, 5)$ be $x^2 + y^2 + 2gx + 2fy + k = 0$. Substituting the points: For $(2,

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$ \frac{14}{3}, 1 $

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(C) Let the equation of the circle passing through $(2, 0), (0, 1)$ and $(4, 5)$ be $x^2 + y^2 + 2gx + 2fy + k = 0$. Substituting the points: For $(2, 0): 4 + 4g + k = 0 \Rightarrow 4g + k = -4$ For $(0, 1): 1 + 2f + k = 0 \Rightarrow 2f + k = -1$ For $(4, 5): 16 + 25 + 8g + 10f + k = 0 \Rightarrow 8g + 10f + k = -41$ Solving these equations: $k = -4 - 4g$ $2f = -1 - k = -1 - (-4 - 4g) = 3 + 4g \Rightarrow f = \frac{3}{2} + 2g$ Substitute into the third equation: $8g + 10(\frac{3}{2} + 2g) + (-4 - 4g) = -41$ $8g + 15 + 20g - 4 - 4g = -41$ $\Rightarrow 24g = -52$ $\Rightarrow g = -\frac{13}{6}$ Then $k = -4 - 4(-\frac{13}{6}) = -4 + \frac{26}{3} = \frac{14}{3}$ And $f = \frac{3}{2} + 2(-\frac{13}{6}) = \frac{3}{2} - \frac{13}{3} = \frac{9-26}{6} = -\frac{17}{6}$ The circle equation is $x^2 + y^2 - \frac{13}{3}x - \frac{17}{3}y + \frac{14}{3} = 0$. Since $(0, c)$ lies on the circle: $0^2 + c^2 - \frac{13}{3}(0) - \frac{17}{3}c + \frac{14}{3} = 0$ $3c^2 - 17c + 14 = 0$ $(3c - 14)(c - 1) = 0$ Thus,$c = \frac{14}{3}$ or $c = 1$.

If the points $(2, 0), (0, 1), (4, 5)$ and $(0, c)$ are concyclic,then $c$ is equal to

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